Let $D\in \mathbb{R}^{n\times n}$ denote a lower triangular matrix. With $\|\cdot\|$ denoting the spectral matrix norm, is there an estimate like
$$ \|D\| \leq C\|D+D^T\|, $$
where $C>0$ is independent of the dimension $n\in\mathbb{N}$ and $D$?
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1$\begingroup$ No. The best constant is $C\sim\ln n$. See this question and the link provided there for background: mathoverflow.net/questions/177198/… $\endgroup$– Christian RemlingCommented Aug 6, 2014 at 16:32
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$\begingroup$ @ChristianRemling I think you could post this as an answer $\endgroup$– Yemon ChoiCommented Aug 6, 2014 at 17:57
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$\begingroup$ @YemonChoi: OK, will do. $\endgroup$– Christian RemlingCommented Aug 6, 2014 at 18:10
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3 Answers
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No. The best constant is $C_n\sim \ln n$. See, for example, this paper. In particular, for the lower bound, see example 3.3 of the paper.
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$\begingroup$ Name of this paper: Bhatia - Pinching, Trimming, Truncating, and Averaging of Matrices. $\endgroup$– LSpiceCommented Jan 20, 2022 at 3:54
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The $\log n$ result mentioned by Christian Remling is a special case of the results in
Kwapień, S.; Pełczyński, A. The main triangle projection in matrix spaces and its applications. Studia Math. 34 1970 43–68.
answered Aug 6, 2014 at 18:30
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The assumption that $D$ is triangular is not necessary (the Schur decomposition may bring a matrix to its triangular form). If the numerical range of D is contained in a sector, then Lemma 3.1 in http://www.tandfonline.com/doi/abs/10.1080/03081087.2014.933219#preview says that there is a C in terms of secant function.
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$\begingroup$ Unless I have misremembered, even if a square matrix $A$ is unitarily conjugate to a triangular matrix $D$, it might not be the case that $A+A^\top$ is unitarily conjugate to $D+D^\top$. Therefore I think that the question as stated really does want to start with the assumption that $D$ is triangular. $\endgroup$ Commented Aug 16, 2014 at 18:43
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$\begingroup$ @Yemon, Yes, you are right. If the transpose $^T$ is replaced with the conjugate transpose $^*$, then my comment is valid. $\endgroup$– M. LinCommented Aug 17, 2014 at 3:48
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$\begingroup$ Name of the linked paper: Zhang - A matrix decomposition and its applications. $\endgroup$– LSpiceCommented Jan 20, 2022 at 3:54